Usually Diazo coupling occurs at para-position unless the para-position is occupied, in which case coupling occurs at ortho-position.

While solving questions I found in both the cases as shown, the coupling occurred at ortho-position of the $\ce{-OH}$ group:

enter image description here

So in case of phenol, is it possible that in all cases the coupling will occur at ortho-position with respect to the $\ce{-OH}$ group in phenol, irrespective of whether the other group has stronger +M Effect than $\ce{-OH}$?

I am not sure but I think this is happening because of Hydrogen bonding as shown below. But in that case it should also happen with the $\ce{-NH2}$ group. Is this the reason or there is some other?

enter image description here

  • 2
    $\begingroup$ If there’s hydroxide ion there, then the phenol will be deprotonated. $\endgroup$ Mar 21, 2019 at 20:17
  • $\begingroup$ @orthocresol True , the Books Says Appreciable amount of Phenol will be converted in phenoxide ion . But Still the question and options all have -OH . $\endgroup$ Mar 21, 2019 at 20:42
  • 2
    $\begingroup$ Don't get so caught up about what the question writes... You put in a phenol into the reaction, it gets converted into phenoxide, that's all there is to it. Just because somebody drew a phenol on a piece of paper doesn't mean that it can't be deprotonated. $\endgroup$ Mar 21, 2019 at 21:01
  • $\begingroup$ I'm not sure what you're trying suggest with that hydrogen bond. This is probably an exergonic reaction, in which case, the Hammond Postulate would suggest an early transition state where the form of the product has little impact on the rate of the reaction... $\endgroup$
    – Zhe
    Mar 21, 2019 at 23:26
  • $\begingroup$ @orthocresol: comparing the influence of steric hindrance, strength of activating resonance, and hydrogen-bonding stability, I concluded that(say the position ortho to the -OH is 1 and the other 2) for -OCH3, position 1 is favoured twice and hydrogen-bonding isn't a factor(so no ambiguity on the answer there), and for -NH2, all factors except steric hindrance favour position 2. Could you explain how the steric hindrance factor alone makes 1 more favorable? And while I appreciate the rigor in Zhe's comment, neither me nor the OP(at the time of posting) require that level of Chem applied here. $\endgroup$
    – harry
    May 4, 2021 at 15:06


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.